Question

y3-3y-52=0 what is value of y

Answer 1

By hit and trial, one of the roots is 4.
therefore:
$f(x) = 0 \\ {y}^{3} - 3y - 52 = 0 \\ (y - 4)( {y}^{2} + 4y + 13) = 0 \\ now \\ (y - 4) = 0 \: \: \: \: or \: \: ( {y}^{2} + 4y + 13) = 0 \\ y = 4 \: \\ ( {y } ^{2} + 4y + 13 = 0) \: has \: no \: real \: solutions.$
If you are satisfied, please rate my answer as the Branliest answer.